Question

If $a\in [-10,0]$, then the probability that the graph of the function $y=x^2+2(a+3)x-(2a+6)$ is strictly above $x$ -axis, is

• A. $\frac{3}{5}$
• B. $\frac{2}{5}$
• C. $\frac{1}{5}$
• D. None of these

Answer 1

Answer: (C)

(i) As $f(x)=y=x^2+2(a+3)x-2a-6>0$
$\forall x\in R$
(ii) $f(x)>0$, coefficient of $x^2$ is $1>0$
(iii) $D<0$
$\therefore 4(a+3{)}^2-4(-2a-6)<0$
$\Rightarrow (a+3{)}^2+2a+6<0$
$\Rightarrow a^2+8a+15<0$
$\Rightarrow (a+3)(a+5)<0$
$\Rightarrow -5<a<-3$
Now, $n(S)=$ length of interval, as $n\in [-10,0]$
$=0-(-10)=10$
$\therefore n(S)=10$
$n(A)=$ length of interval, when $-5<a<-3$
$=-3-(-5)=2$
$\therefore$ Required probability $=\frac{n(A)}{n(S)}$
$=\frac{2}{10}=\frac{1}{5}$
Alternative Solution :
As $f(x)=y>0$ ,br> $\Rightarrow a>0$ and $D<0$
$\Rightarrow a^2+8a+15<0$
$\Rightarrow -5<a<-3$ (To be use for $n(A))$
$\therefore$ Required probability $=\frac{n(A)}{n(S)}$
$=\frac{\underset{-5}{\overset{-3}{\int }}dx}{\underset{-10}{\overset{0}{\int }}dx}=\frac{2}{10}=\frac{1}{5}$